The two wires A and B of equal cross-section but of different materials are joined together. The ratio of Young's modulus of wire A and wire B is 20/11 . When the joined wire is kept under certain tension the elongations in the wires A and B are equal. If the length of wire A is 2.2 m, then the length of wire B is _______ m.

Step-by-step solution

Given that the wires have equal cross-sectional area A and are under the same tension F , the elongation L is given by: L = F L A Y Since the elongations in both wires are equal ( L_A = L_B ): F L_A A Y_A = F L_B A Y_B L_A Y_A = L_B Y_B L_B = L_A Y_B Y_A Substituting the given values L_A = 2.2 m and Y_A Y_B = 20 11 : L_B = 2.2 11 20 L_B = 24.2 20 = 1.21 m Answer: 1.21